How to reasonably estimate the probability of your father being exactly 12222 days older than you?

On 2024/05/03, my brother is 10,000 days of age, while my father will be 22,222 days old. To tell them as a fun-fact, I would like to grab a feel of how special this is. I have no access to datasets that could provide me a distribution of the likelihood of becoming a parent at n days of age. It doesn't matter whether this child is the father's first child.

(Off-topic: I asked chat gpt for inspiration on how to celebrate this event, and to my surprise it was quite creative as long as it didn't have to organize it itself.)

Kind regards from Belgium

• "It doesn't matter whether this child is the fathers first child." In fact, it does not matter at all if he is the fathers ot "his" child. Forecasts of growth in population base only on momens quota. BTW, while you are at it, my father was 21.850 days older than me (due to 2nd WW) -- how likely is that? Feb 7 at 0:45
• I think we have to try to define what counts as "special". Would it also be similarly "special" if on that date, your dad were 15000/16000/17000/18000/19000/20000/21000 etc. days old? What about 15555/16666/17777/etc.? What about 23456? Feb 7 at 2:03
• The probability of being exactly 12,222 days old is probably quite low, but not significantly higher or lower than being exactly 12,221 or 12,223 days old. There's also the question of whether you want the absolutely probability P(12222 days old) or some conditional probability P(12222 days old | age between X and Y days) for some age range X-Y days old. (The narrower the range, the higher the probability, assuming 12222 is in [X, ]. If it isn't, of course, the probability is zero.) Feb 7 at 15:10
• @user182601: Reminds me of that quote from Richard Feynman: “You know, the most amazing thing happened to me tonight... I saw a car with the license plate ARW 357. Can you imagine? Of all the millions of license plates in the state, what was the chance that I would see that particular one tonight? Amazing!” Feb 8 at 20:38
• @MichaelSeifert: These days, I imagine it would be pretty low; that car is probably from the 1950s (maybe 1960s)! Feb 8 at 23:24

Oops, you do say that you're from Belgium (I somehow missed that at first), but I'm going to use statistics from the U.S., because that's what I found first.

If your brother turns $$10\,000$$ days old on $$2024$$-$$05$$-$$03$$, then they were born in December $$1996$$. At that time, fathers' ages were approximately normally distributed with a mean of about $$\mu \approx 30.0$$ and a standard deviation of about $$\sigma \approx 6.7$$. Your father was about $$33.5$$ years old, so about $$z \approx 3.5/6.7 \approx 0.52$$ standard deviations above the mean.

The probability density function (PDF) of the standard normal distribution is

$$f(z) = \frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}}$$

Here, that evaluates to $$f(0.52) \approx 0.35$$. One day is about $$\frac{1}{6.7\cdot365} \approx 0.00041$$ standard deviations, so the probability of falling in a one-day slice $$3.5$$ years above the mean is about $$0.00041 \times 0.35 \approx 0.00014$$, or about one chance in $$7000$$. (I kept some extra digits around till the end.)

I wouldn't trust that figure very far, but this gives an idea of how to ball park it. It's not very far off Robert Israel's result. Keep in mind, though, that there are probably quite a few numbers you would consider as remarkable as $$12\,222$$.

ETA (2024-02-09). Bringing a caveat from the comments into this answer: ajd138 from the comments suggested that the normal approximation wasn't appropriate for the father's age at the child's birth. That's clearly true at the extremes, but in this case the age is less than a standard deviation above the mean, and my guess is that the normal approximation is not bad in the neighborhood of the mean.

This answer was intended to show a method for using a parametric distribution—the normal distribution, in this case—for obtaining a probability. If such a distribution is not available, one can use the method in Dan's answer and simply find a sufficiently fine histogram for the probability. However, if someone has an idea for a better parametric distribution, I'm all ears!

Note also that Amir's answer raises an interesting issue with respect to the father's probability of surviving to see their child's $$10\,000$$th day. The impact of conditioning on the father's survival is an interesting one, although I think the question's interpretation doesn't obviously rest on that condition.

• If you assume that the father-son age difference is between 20 and 40 years, then there are 8 possible day counts that are "remarkable" in having the last 4 digits the same: 7777, 8888, 9999, 10000, 11111, 12222, 13333, and 14444.
– Dan
Feb 7 at 17:40
• I don't think a normal distribution is appropriate in this case Feb 7 at 17:54
• @ajd138: Certainly in the extremes it is not. Near the mean, I don't think it's a bad approximation, and I submit it's better than a uniform distribution across some interval. Do you have a better parametric one for the data I adduced, by any chance? ETA: Oops, I switched sources at some point; I don't think the one I linked to actually has a breakdown by age. Sorry! Still interested to hear if you have a good alternative characterization, though. Feb 7 at 18:02
• @Dan: There might be others—$12\,221, 12\,321, 11\,211, \ldots$. It's hard to say exactly what the OP might find equally remarkable, but there are surely quite a few, as I say! Feb 7 at 18:04
• I'm really flattered to see so much solid responses(!) thanks to all of you! The answers indeed converge to an order of magnitude of 1/7000, therefore suggesting that if I would consider my brother as a normal representative random sampled Belgian, then 1655 Belgians would normally already share this unique special case^^. So unworthy for instance to bother a local press instance to cram this into a little article^^. Feb 8 at 1:39

12,222 days = 33 years (and about 5½ months)

I haven't been able to find information specific to Belgium, but if you assume that the age distribution of fathers is similar to Norway, then the probability of your dad being 33 when you're born is approximately 5%.

If you assume that all birth dates between your dad's 33rd and 34th birthdays are equally probable, then the probability that he's exactly 12,222 days older than you works out to $$\frac{0.05}{365} \approx 0.000137$$, or about 1 in 7300.

• ... and so about $1$ in $7300$ just like @BrianTung's answer Feb 7 at 1:10
• I'm really flattered to see so much solid responses(!) thanks to all of you! The answers indeed converge to an order of magnitude of 1/7000, therefore suggesting that if I would consider my brother as a normal representative random sampled Belgian, then 1655 Belgians would normally already share this unique special case^^. So unworthy for instance to bother a local press instance to cram this into a little article^^. Feb 8 at 1:42

In Western societies, males typically have children from, let's say, age $$20$$ to $$45$$ years, or about $$7300$$ to $$16400$$ days. I haven't looked up detailed statistics, and they would in any case differ from country to country and from year to year. That's a span of about $$9100$$ days, and as a very rough estimate any of those $$9100$$ days would have probability about $$1/9100$$ of being the difference between your father's age and yours.

• I'm really flattered to see so much solid responses(!) thanks to all of you! The answers indeed converge to an order of magnitude of 1/7000 or indeed a 1/9100 underbound based on a uniform distribution baseline. This suggests that if I would consider my brother as a normal representative random sampled Belgian, then 1655 Belgians would normally already share this unique special case^^. So unworthy for instance to bother a local press instance to cram this into a little article^^. Feb 8 at 1:46
• "this unique special case" I think it's important that you identify just what the unique special case actually is. Is it that $12,222$ is an interesting number because of all the $2$s. Had the been $13,333$ days apart would that have been boring and uninteresting? What if they were born $12,345$ days apart? What if they were born, $10,529$ days apart? The probability of that happening is exactly the same as being born $12,222$ days apart. Feb 25 at 8:31

For computing the probability, it is not enough to only focus on the father's paternal age. We also need to consider that both father and brother survive until 05-03-2024.

As described below in detail, what computed in the other three answers is the probabilty of the following event:

$$B$$: The father has a child 10,000 before 05-03-2024.

and not the event given in the OP:

$$A$$: On 05-03-2024, the father is 12222 days older than the brother (his son).

One can see tha A happens only if the following event also occurs:

$$\color{blue}{C}$$: The father and his son survive until 05-03-2024.

That is

$$A=B \cap \color{blue}{C}.$$

Thus, the probability of that on 05-03-2024 the father is 12222 days older than the brother, event $$A$$, is

$$\mathbb P(A)=\mathbb P(B \cap\color{blue}{C})=\color{blue}{\mathbb P \left ( \text{Brother's age} \ge 10000 \right ) \times \\ \mathbb P \left ( \text{Father's age} \ge 22222 \, \big | \, \text{Father's age} \ge 12222 \right )} \times \\ \mathbb P \left ( \text{Father's paternal age} \in [12222,12222+1] \right ).$$

Above, we hiddenly assume that 1) the father was alive 10000 days ago, 2) the residual life is not affected by the paternal time, and 3) the reference time for calculation of the probability is a time instant before the father becomes 18 years old; see case A.1 in the first part below on reference time.

If the lifespan (LS) of a Belgian man follows pdf $$f_{LS}$$ with cdf $$F_{LS}$$, and the paternal age (PA) of a Belgian man follows pdf $$f_{PA}$$ with cdf $$F_{PA}$$; then the probability is approximately:

$$\fbox{\color{blue}{\left (1-F_{LS} (10000) \right )\frac{ 1-F_{LS}(22222) }{ 1-F_{LS} ( 12222 ) } } f_{PA} ( 12222 ) }$$

in which the following approximation is used: $$F_{PA} (12222+1 )-F_{PA} (12222)\approx f_{PA} (12222 ).$$

Computation of probability when both LS and PA distributions are normal

If the lifespan of a Belgian man follows $$N(\mu_1,\sigma_1^2)$$, and the paternal age of a Belgian man follows $$N(\mu_2,\sigma_2^2)$$; the probability is approximately:

$$\color{blue}{\left (1-\Phi \left ( \frac{10000-\mu_1}{\sigma_1} \right ) \right ) \times \\ \frac{\left (1-\Phi \left ( \frac{22222-\mu_1}{\sigma_1} \right ) \right )}{1-\Phi \left ( \frac{12222-\mu_1}{\sigma_1} \right) }} \times \\ \frac{1}{\sqrt{2\pi}\sigma_2} e^{-\frac{(12222-\mu_2)^2}{2\sigma_2^2}} .$$

For

$$\mu_2 \approx 30.0×365, \sigma_2 \approx 6.7×365,$$

it can be significantly smaller than about 1 in 7000, obtained in other answers depending on the mean (life expectancy) $$\mu_1$$ and standard deviation $$\sigma_1$$ of the lifespan.

In fact, the probability can be written as

$$\color{blue}{k }\times \frac{1}{7000}$$

where $$k=\left (1-\Phi \left ( \frac{10000-\mu_1}{\sigma_1} \right ) \right ) \times \frac{\left (1-\Phi \left ( \frac{22222-\mu_1}{\sigma_1} \right ) \right )}{1-\Phi \left ( \frac{12222-\mu_1}{\sigma_1} \right) }.$$

Let $$\sigma_1=\sigma_2 \approx 6.7×365$$ and consider the following scenarios:

for $$\mu_1 \approx 50.0×365$$, $$\, \color{blue}{k=0.05251}$$

for $$\mu_1 \approx 60.0×365$$, $$\, \color{blue}{k= 0.44764}$$

for $$\mu_1 \approx 70.0×365$$, $$\, \color{blue}{k=0.91322}$$

for $$\mu_1 \approx 80.0×365$$, $$\, \color{blue}{k=0.99784}.$$

Hence, when the life expectancy is greater than 80 years old, which is the case for Belgian men according to this site, then $$k \approx 1$$ (under the normality assumption, which is strongly criticized below; see the second part below on distributions).

Reference time for calculation of probability

To calculate the probability of a temporal event, we must fix the time at which we calculate the probability, called here reference time. The level of uncertainty of the event varies over the time. Different cases are considered below (the adverb ago in the following stands for the days before the specific date of 05-03-2024, not current date).

A- The reference time for calculation of the probability is before the birth of the brother (so we need to take into account the survival of both too). Here, we have two sub-cases:

A.1. The reference time is before $$22222-18 \times 365$$ days ago (we have done our analysis above based on this assumption).

A.2. The reference time is between $$22222-18 \times 365$$ days ago and $$22222-10000$$ days ago.

In A.2, as the father gets older, the distribution of the paternal age may be affected and needs to be updated using a conditional distribution.

In both A.1 and A.2, we assume that the father's paternal age is always greater than 18.

B- The reference time for calculation of the probability is after the birth of the brother and before the specific date of 05-03-2024.

In this case, we don't need to take into account that the father becomes a parent 10000 days ago, as he is a father now.

However, the event is still uncertain. In this period, the probability only depends on that both father and brother survive until the specific date in the remaining days.

C- The reference time for calculation of the probability is after the specific date of 05-03-2024. In this case, the event becomes fully certain, and the probability of occurrence of the event is nothing but 1.

Remark: One may think we can ignore the surviving part of the event (considering it as here-and now, occurring with probability 1) and only focus on the birth part as wait-and- see, because when the father and brother are informed about the event they are required to be alive; while based on the above described cases, you can clearly see that in this reasoning two different reference times for calculation of the probabilities of the birth and surviving parts of the event are hiddenly considered, which is not correct.

Distributions of lifespan and paternal age

As pointed in a comment below by @Nuclear Hoagie, the normal distribution may be problematically narrow for the lifespan, as it puts 95% of people within 13.4 years of the average lifespan. For $$\mu_ 1=50$$, it means almost no one lives to even 65, and for $$\mu_ 1=80$$, almost all pass 65 years old. Hence, more reasonable distributions for lifespan $$f_{LS}$$ and the paternal age $$f_{PA}$$ need to be considered.

This paper, on a data set from Swiss, shows that proportional hazard (PH) Weibull and Gompertz model and skew-normal distribution are the best for the lifespan distribution, shown below for men and women:

• I'm not quite sure how this gets to an answer an order of magnitude different than the others, it seems the main difference is including the probability both people survive long enough (in addition to the probability their birthdays are at the right interval to begin with). But that would imply that only 1 in 10 father-son pairs survive to the son's 28th birthday, which surely cannot be the case. Feb 7 at 15:13
• I grant the point. What is $k$ in this case? Feb 7 at 18:08
• @BrianTung Thank you! I just added $k$ for different scenarios. It depends on the mean and standard deviation of Belgian men.
– Amir
Feb 7 at 18:13
• I find the normal distribution of lifespans to be problematically narrow - it puts 95% of people within 13.4 years of the average lifespan. At u1=50, it means almost no one lives to even 65, and at u1=80, it means practically everyone lives past 65, but those aren't very good models of population. A Belgian life expectancy of 80 years doesn't indicate that nearly 99.9% of Belgians reach their 60th birthday, which is what the normal distribution using an SD of 6.7 years implies. Feb 7 at 18:57
• @Amir: Although—I wonder if OP would have found it any less remarkable in the unfortunate case where his father did not survive to see his $22\,222$nd day. Feb 7 at 19:16